Is the longest Weyl transformation of a product group the pair of the longests? More precisely, if $G_1,G_2$ are two Lie groups and $T_1,T_2$ are maximal tori respectively, then $W(G_1\times G_2)\cong W(G_1)\times W(G_2)$. 
The pair of longest elements in $W(G_1)\times W(G_2)$ is mapped to some element in $W(G_1\times G_2)$ and I would expect this is the longest. 
Is this a true statement and can someone show a proof? 
Thanks in advance. 
 A: If $\Phi_1$ and $\Phi_2$ denote the root systems of $G_1$ and $G_2$ w.r.t $T_1$ and $T_2$, then $\Phi = \Phi_1 \sqcup \Phi_2$ is the root system of $G_1 \times G_2$.
Let us write $W_i = W(G_i)$ and set $W = W_1 \times W_2$.
Now, we only can speak of longest elements after having fixed sets of simple roots $\Delta_i \subseteq \Phi_i$ and then $\Delta = \Delta_1 \sqcup \Delta_2$ is a set of simple roots of $\Phi$.
Note that the longest element of a Weyl group is uniquely determined by mapping all positive roots to negative ones (in fact the length of a Weyl group element is just the number of positive roots that become negative under this element). 
So if $w_i \in W_i$ denotes the longest element of $W_i$ w.r.t. $\Delta_i$ we find that $w_1$ maps every positive root in $\Phi_1$ to a negative one and fixes every root in $\Phi_2$ and similarly for the longest element $w_2$ of $W_2$ w.r.t. $\Delta_2$. Thus $w = w_1 w_2 = w_2 w_1$ maps every positive root in $\Phi$ to a negative one and thus is the longest element.
